Part A A uniform solid disk of radius 1.60 m and mass 2.30 kg rolls without slip

Question

Part A

A uniform solid disk of radius 1.60 m and mass 2.30 kg rolls without slipping to the bottom of an inclined plane. If the angular velocity of the disk is5.58 rad/s at the bottom, what is the height of the inclined plane?

A uniform solid disk of radius 1.60 m and mass 2.30 kg rolls without slipping to the bottom of an inclined plane. If the angular velocity of the disk is 5.58 rad/s at the bottom, what is the height of the inclined plane?

6.1 m

Problem 3

Force F =14j^N is exerted on a particle at r =(3i^+5j^)m.

Part A

What is the torque on the particle about the origin?

Express your answer using two significant figures. Enter coordinates numerically separated by commas. x,y,z =

6.1 m

Explanation / Answer

mass of the solid sphere, m=2.3 kg

radius of the solid sphere, r=1.6m

angular velocity w=5.58 rad/sec

for solid disk moment of inertia I=1/2*m*r^2

by using law of conservation of energy,

E_top=E_bottom

P.E=K.E+R.E

m*g*h=1/2*m*v^2+1/2*I*w^2

m*g*h=1/2*m*v^2+1/2*(1/2*m*r^2)*w^2

g*h=1/2*r^2*w^2+1/2*(1/2*r^2*w^2)

h=(3/4g)*(r*w)^2

h=(3/(4*9.8))*(1.6*5.58)^2

====> h=6.1 m

the height of the inclined plane is h=6.1 m

3)

position of particle r=(3i+5j) in m

force F=-14 j in Joule

torque T=rXF

T=(3i+5j)X(-14 j)

T=-42k in N.m

there fore Tx=0, Ty=0, Tz=-42 N.m

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